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RSB [31]
3 years ago
8

Fernell and Dabney shared the driving

Mathematics
1 answer:
Agata [3.3K]3 years ago
5 0

Answer:

15 hours

Step-by-step explanation:

Given:

Fernell's speed v_F= 50 mph

Dabney's speed v_D= 64 mph

Denote:

D_F - distance covered by Fernell

D_D - distance covered by Dabney

t_F - Fernell's time

t_D - Dabney's time

1. If Fernell drove for 3 hours longer than Dabney, then his time is 3 hours more than Dabney's time and

t_F=t_D+3

2. If Fernell covered 18 miles less than Dabney, then

D_D-D_F=18

Use formula D=t\cdot v

D_D=v_D\cdot t_D\Rightarrow D_D=64\cdot t_D\\ \\D_F=v_F\cdot t_F\Rightarrow D_F=50\cdot (t_D+3)

Subtract from the first equation the second equation and equate it to 18:

64t_D-50(t_D+3)=18\\ \\64t_D-50t_D-150=18\\ \\14t_D=168\\ \\t_D=12\ \text{hours}

t_F=t_D+3=12+3=15\ \text{hours}

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Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

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The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 36 hours and a standard deviation of 5.5 hours.

This means that \mu = 36, \sigma = 5.5

a. What can you say about the shape of the distribution of the sample mean?

By the Central Limit Theorem, it is approximately normal.

b. What is the standard error of the distribution of the sample mean? (Round your answer to 4 decimal places.)

Sample of 9 means that n = 9. So

s = \frac{\sigma}{\sqrt{n}} = \frac{5.5}{\sqrt{9}} = 1.8333

The standard error of the distribution of the sample mean is 1.8333.

c. What proportion of the samples will have a mean useful life of more than 38 hours?

This is 1 subtracted by the pvalue of Z when X = 38. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{38 - 36}{1.8333}

Z = 1.09

Z = 1.09 has a pvalue of 0.8621

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d. What proportion of the sample will have a mean useful life greater than 34.5 hours?

This is 1 subtracted by the pvalue of Z when X = 34.5. So

Z = \frac{X - \mu}{s}

Z = \frac{34.5 - 36}{1.8333}

Z = -0.82

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pvalue of Z when X = 38 subtracted by the pvalue of Z when X = 34.5. So

0.8621 - 0.2061 = 0.656

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